Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/84

${\frac {1}{x}}=y,\qquad \operatorname {d} x=-{\frac {\operatorname {d} y}{y^{2}}}\,;$

et par conséquent

$e^{-{\frac {1}{x}}}\operatorname {d} x=-{\frac {1}{y^{2}}}e^{-y}\operatorname {d} y=-y^{-2}.e^{-y}\operatorname {d} y.$

Posant alors

$P=y^{-2},\qquad Q=e^{-y},$

nous aurons d’une part

${\begin{alignedat}{1}{\frac {\operatorname {d} P}{\operatorname {d} y}}=&-2y^{-3},\\\\{\frac {\operatorname {d} ^{2}P}{\operatorname {d} y^{2}}}=&+2.3y^{-4},\\\\{\frac {\operatorname {d} ^{3}P}{\operatorname {d} y^{3}}}=&-2.3.4y^{-5},\\\ldots &\ldots \ldots \ldots \ldots \\{\frac {\operatorname {d} ^{\mu }P}{\operatorname {d} y^{\mu }}}=&(-1)^{\mu }.2^{\mu |1}y^{-(\mu +2)}.\qquad \mathrm {(10)} \end{alignedat}}$

D’une autre part, nous aurons

${\begin{alignedat}{1}{\frac {\operatorname {d} Q}{\operatorname {d} y}}=&-e^{-y},\\\\{\frac {\operatorname {d} ^{2}Q}{\operatorname {d} y^{2}}}=&+e^{-y},\\\\{\frac {\operatorname {d} ^{3}Q}{\operatorname {d} y^{3}}}=&-e^{-y},\\\ldots &\ldots \ldots \ldots \ldots \\{\frac {\operatorname {d} ^{\mu }Q}{\operatorname {d} y^{\mu }}}=&(-1)^{\mu }.e^{-y}\,;\qquad \mathrm {(11)} \end{alignedat}}$